Variables - Wrong approach.

[gmatrant]

A town's oldest inhabitant is x years older than the sum of the ages of the Lee triplets. If the oldest inhabitant is now J years old, how old will one of the triplets be in 20 years?

Ans: [spoiler](J-X+60)/3[/spoiler]

My approach - Can you please let me know where I am going wrong
Let L be the sum of the ages of the Lee triples (all three of them)
J = x + L ------- (i){based on the first line in the problem}
20 years from now
J + 20 = x + L + 60 {adding 20 for each of the lee brothers}
L = J -40 -X
Age of one of the triplets can be found by dividing L by 3.
so L/3 ==> (J-40-X)/3.

But this is not the right answer. Can you please tell me where I am going wrong?

[neelgandham]

Before I point out the mistakes. Here is how I solved the problem.

J = age of the town's oldest inhabitant
L = age of each of Lee's triplets

J = X + 3L
(J-X)/3 = L

Question asked :how old will one of the triplets be in 20 years? can be rephrased to 'What is the value of L + 20?'

L + 20 = ((J-X)/3)+20 = (J-X+60)/3.

[neelgandham]

20 years from now
J + 20 = x + L + 60 {adding 20 for each of the lee brothers}
Can you please tell me where I am going wrong?

This is the mistake. The question states A town's oldest inhabitant is x years older than the sum of the ages of the Lee triplets. and NOT In 20 years, A town's oldest inhabitant is x years older than the sum of the ages of the Lee triplets.

[user123321]

given J = 3y + x
where y is age of child.
so 3y = J-x
y = (J-x)/3
after 20 years child's age will be y+20
=> y+20 = (J-x)/3 + 20
= (J-x+60)/3

user123321

[GMATGuruNY]

A town's oldest inhabitant is x years older than the sum of the ages of the Lee triplets. If the oldest inhabitant is now J years old, how old will one of the triplets be in 20 years?

A. (J-50)/3

B. (3(J+20))/x

C. (J+x-50)/3

D. (J-x +60)/3

E. (J+x-20)/3

Let the age of each triplet = 2.
The sum of the triplets' ages = 3*2 = 6.
Let x=5.
J = the age of the oldest inhabitant = 6+5 = 11.
In 20 years, the age of each triplet = 2+20 = 22. This is our target.

Now we plug x=5 and j=11 into the answers to see which yields our target of 22.

Only answer choice D works:
(j-x+60)/3 = (11-5+60)/3 = 66/3 = 22.

The correct answer is D.

[chieftang]

Mitch's quote shows 5 answer choices, but when I look at the original question in the thread, there are no answer choices. So, given the answer choices shown by Mitch, the plug-in method is definitely useful and is a good case for being sharp at mental math.

With no answer choices, this is how I went about it. And I think both methods should be practiced. There are many ways to set these things up, but for me, the best way is to make sure I always have a variable for the quantity I'm trying to determine. It sounds obvious, but I think it's crucial. Give it a shot, and practice it on other problems until it makes sense: Always have a variable for the quantity you're trying to determine.

OK. What am I trying to determine here? Answer: the age of a single triplet in 20 years.

So, I let T = age of a triplet in 20 years.

What other quantities are there in this problem? Answer: The age of the oldest inhabitant now, J.

How do these quantities interact with each other? Answer: The age of the oldest inhabitant (J) is x + SUM(age of three triplets today).

So, if T is the age of one triplet 20 years from today, then SUM(age of three triplets today) = 3(T - 20). Therefore: J = 3(T - 20) + x

And solving for T we get:
J = 3T - 60 + x
3T = J + 60 - x
or:
T = (J + 60 - x)/3